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EE130/230A Discussion 2

EE130/230A Discussion 2. Peng Zheng. Electron and Hole Concentrations. Silicon doped with 10 16 cm -3 phosphorus atoms, at room temperature ( T = 300 K). n = N D = 10 16 cm -3 , p = 10 20 /10 16 = 10 4 cm -3

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EE130/230A Discussion 2

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  1. EE130/230A Discussion 2 PengZheng

  2. Electron and Hole Concentrations • Silicon doped with 1016 cm-3 phosphorus atoms, at room temperature (T = 300 K). n = ND = 1016 cm-3, p = 1020/1016 = 104 cm-3 • Silicon doped with 1016 cm-3 phosphorus atoms and 1018 cm-3 boron atoms, at room temperature. p =NA - ND = 1018 cm-3, n = 1020/1018 = 102 cm-3 For a compensated semiconductor, i.e. one that has dopants of both types, it is the NET dopant concentration that determines the concentration of the majority carrier. Use np = ni2 to calculate concentration of the minority carrier.

  3. Electron and Hole Concentrations • Silicon doped with 1016 cm-3 phosphorus atoms and 1018 cm-3 boron atoms, at T = 1000 K NA = 1018 cm-3, ND = 1016 cm-3 ni = 1018 cm-3 at T = 1000 K • If ni is comparable to the net dopant concentration. Then the equations on Slide 17 of Lecture 2 must be used to calculate the carrier concentrations accurately. Note, np = ni2 is true at thermal equilibrium.

  4. n(Nc, Ec) and p(Nv, Ev)

  5. n(ni, Ei) and p(ni, Ei) • In an intrinsic semiconductor, n = p = ni and EF = Ei

  6. n-type Material R. F. Pierret, Semiconductor Device Fundamentals, Figure 2.16 Energy band diagram Density of States Carrier distributions Probability of occupancy EE130/230A Fall 2013 Lecture 3, Slide 6

  7. p-type Material R. F. Pierret, Semiconductor Device Fundamentals, Figure 2.16 Energy band diagram Density of States Carrier distributions Probability of occupancy EE130/230A Fall 2013 Lecture 3, Slide 7

  8. Fermi level applets • http://jas.eng.buffalo.edu/

  9. Energy band diagram • Consider a Si sample maintained under equilibrium conditions, doped with Phosphorus to a concentration 1017 cm-3. For T = 300K, indicate the values of (Ec – EF) and (EF – Ei) in the energy band diagram. • Since ni= 1010 cm-3 and n = nie(EF-Ei)/kT: • EF – Ei= kT(ln107) = 7 ∙ kT(ln10) = 7 ∙ 60 meV = 0.42 eV • The intrinsic Fermi level is located slightly below midgap: • Ec– Ei= Ec– [(Ec+Ev)/2 + (kT/2)∙ln(Nv/Nc)] • = (Ec– Ev)/2 – (kT/2)∙ln(Nv/Nc) = 0.56 eV + 0.006 eV = 0.566 eV • Hence Ec– EF = (Ec – Ei) – (EF – Ei) = 0.566 – 0.42 = 0.146 eV

  10. Energy band diagram • For T = 1200K, indicate the values of (Ec – EF) and (EF – Ei). Remember that Nc and Nv are temperature dependent. Also, EG is dependent on temperature: for silicon, EG = 1.205  2.8×10-4(T) for T> 300K. At T = 1200K, the Si band gap EG = 1.2  2.8×10-4 (T) = 0.87 eV The conduction-band and valence-band effective densities of states Nc and Nv each have T3/2 dependence, so their product has T3 dependence. (The ratio Nv/Nc does not change with temperature, assuming that the carrier effective masses are independent of temperature.) Therefore, when the temperature is increased by a factor of 4 (from 300K to 1200K), NcNv is increased by a factor of 64.  Intrinsic Semiconductor: EF = Ei

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